A parallelogram is constructed on the vectors ${\bf{\vec a}} = 3\vec \alpha - {\rm{\vec \beta }},\;{\bf{\vec b}} = {\rm{\vec \alpha }} + 3{\rm{\vec \beta }}$, if $\left| {{\rm{\vec \alpha }}} \right| = \left| {{\rm{\vec \beta }}} \right| = 2$ and angle between ${{\rm{\vec \alpha }}}$ and ${\rm{\vec \beta }}\;{\rm{is}}\frac{{\rm{\pi }}}{3},$ then length of diagonal of the parallelogram is

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

1 of 1

Application +4 | -1

A parallelogram is constructed on the vectors

\mathbf{\vec a} = 3\vec \alpha - \mathrm{\vec \beta },\;\mathbf{\vec b} = \mathrm{\vec \alpha } + 3\mathrm{\vec \beta }

, if

\left| {\mathrm{\vec \alpha }} \right| = \left| {\mathrm{\vec \beta }} \right| = 2

and angle between

{\mathrm{\vec \alpha }}

and

\mathrm{\vec \beta }\;\mathrm{is}\frac{\mathrm{\pi }}{3},

then length of diagonal of the parallelogram is

View answer and explanation

Correct option: B

4\sqrt 3

Option A (

4\sqrt{7}

) is correct because it is the length of one of the diagonals of the parallelogram, calculated using vector addition and the given magnitudes and angle.

More Questions from DOT PRODUCT OF VECTORS

Practice All Chapter MCQs →

The distance between the planes given by

\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

and

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

is A vector which makes equal angles with the vectors

\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right)

, and

{\hat j}

, is If vector

\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and vector

\mathbf{b}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}

, then

\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}}

is equal to The angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

\frac{{5\pi }}{6}

and the projection of

{\mathbf{\vec a}}

in the direction of

{\mathbf{\vec b}}

\frac{{ - 6}}{{\sqrt 3 }}

then

\left| {\mathbf{\vec a}} \right|

is equal to A velocity

\frac{1}{4} \mathrm{~m} / \mathrm{s}

is resolved into two components along

O A

and

O B

making angles

30^{\circ}

and

45^{\circ}

respectively with the given velocity. Then the component along

O B

is Find the equation of the perpendicular drown from the origin to the plane

2x + 4y - 5z = 10